HDU 5412 CRB and Queries 动态整体二分

Problem Description

There are N boys in CodeLand.
Boy i has his coding skill Ai.
CRB wants to know who has the suitable coding skill.
So you should treat the following two types of queries.
Query 1: 1 l v
The coding skill of Boy l has changed to v.
Query 2: 2 l r k
This is a report query which asks the k-th smallest value of coding skill between Boy l and Boy r(both inclusive).

Input

There are multiple test cases.
The first line contains a single integer N.
Next line contains N space separated integers A1, A2, …, AN, where Ai denotes initial coding skill of Boy i.
Next line contains a single integer Q representing the number of queries.
Next Q lines contain queries which can be any of the two types.
1 ≤ N, Q ≤ 105
1 ≤ Ai, v ≤ 109
1 ≤ l ≤ r ≤ N
1 ≤ k ≤ r – l + 1

 

Output

For each query of type 2, output a single integer corresponding to the answer in a single line.

 

Sample Input

5
1 2 3 4 5
3
2 2 4 2
1 3 6
2 2 4 2

Sample Output

3 4

 

题解:

按时间顺序加入,所有的修改拆成删除和添加两个操作,分别对应树状数组中的加减.

维护时间顺序即可

 #include <algorithm>
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 using namespace std;
 const int N=;
 int gi(){
     int str=;char ch=getchar();
     while(ch>'' || ch<'')ch=getchar();
     while(ch>='' && ch<='')str=(str<<)+(str<<)+ch-,ch=getchar();
     return str;
 }
 int n,m,a[N],top=,ans[N];
 struct node{
     int x,y,cnt,k,id;
 }t[N*],q1[N*],q2[N*];
 int Tree[N*];
 void add(int sta,int ad){
     for(int i=sta;i<=n;i+=(i&(-i)))Tree[i]+=ad;
 }
 int getsum(int sta){
     int sum=;
     for(int i=sta;i>=;i-=(i&(-i)))sum+=Tree[i];
     return sum;
 }
 int l1=,l2=;
 void count(int ll,int rr,int dl,int dr)
     {
         for(int i=ll;i<=rr;i++)
             {
                 if(t[i].k)
                     t[i].cnt=getsum(t[i].y)-getsum(t[i].x-);
                 else if(t[i].y<=dr)
                     add(t[i].x,t[i].cnt);
             }
         for(int i=ll;i<=rr;i++)
             if(!t[i].k && t[i].y<=dr)
                 add(t[i].x,-t[i].cnt);
         l1=;l2=;
         for(int i=ll;i<=rr;i++)
             {
                 if(t[i].k)
                     {
                         if(t[i].cnt>=t[i].k)
                             q1[++l1]=t[i];
                         else
                             t[i].k-=t[i].cnt,q2[++l2]=t[i];
                     }
                 else
                     {
                         if(t[i].y<=dr)
                             q1[++l1]=t[i];
                         else
                             q2[++l2]=t[i];
                     }
             }
         int now=ll-;
         for(int i=;i<=l1;i++)
             t[++now]=q1[i];
         for(int i=;i<=l2;i++)
             t[++now]=q2[i];
    }
 void div(int ll,int rr,int dl,int dr)
     {
         if(dl==dr)
             {
                 for(int i=ll;i<=rr;i++)
                     if(t[i].k)ans[t[i].id]=dl;
                 return ;
             }
         int mid=(dl+dr)>>;
         count(ll,rr,dl,mid);
         int to=l1;
         if(to)
             div(ll,ll+to-,dl,mid);
         if(to<=rr-ll)
             div(ll+to,rr,mid+,dr);
     }
 void Clear(){
     top=;
     memset(ans,,sizeof(ans));
     memset(t,,sizeof(t));
 }
 void work()
     {
         Clear();
        for(int i=;i<=n;i++)
         {
             a[i]=gi();
             t[++top]=((node){i,a[i],,,});
         }
         m=gi();
        int flag,x,y,z;
       for(int i=;i<=m;i++)
         {
             flag=gi();
             if(flag==)
                 {
                     x=gi();y=gi();
                     t[++top]=((node){x,a[x],-,,});
                     t[++top]=((node){x,y,,,});
                     a[x]=y;
                 }
             else
                 {
                     x=gi();y=gi();z=gi();
                     t[++top]=(node){x,y,,z,i};
                 }
         }
        div(,top,,1e9);
        for(int i=;i<=m;i++)if(ans[i])printf("%d\n",ans[i]);
     }
 int main()
 {
     while(~scanf("%d",&n))
         work();
     return ;
 }

还有主席树维护的树状数组，但是MLE,这种题目如果数据小可以用

 #include <algorithm>
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 using namespace std;
 const int N=;
 int n,m,a[N],b[N<<];
 struct Ques{
     int flag,x,y,k;
 }q[N];
 int num=,bel[N];
 int getrank(int x){
     int l=,r=num,mid;
     while(l<=r)
         {
             mid=(l+r)>>;
             if(b[mid]==x)return mid;
             if(x>b[mid])l=mid+;
             else r=mid-;
         }
 }
 struct Tree{
     int ls,rs,cnt;
 }t[N*];
 int root[N],tot=;
 void add(int &rt,int l,int r,int val,int ad)
     {
         t[++tot]=t[rt];rt=tot;t[rt].cnt+=ad;
         if(l==r)return ;
         int mid=(l+r)>>;
         if(val<=mid)add(t[rt].ls,l,mid,val,ad);
         else add(t[rt].rs,mid+,r,val,ad);
     }
 void updata(int sta,int val,int ad)
     {
         for(int i=sta;i<=n;i+=(i&(-i)))add(root[i],,num,val,ad);
     }
 int nl[N],nr[N];
 int query(int ll,int rr,int rank)
     {
         int l=,r=num,mid;
         int lenl=,lenr=,sl,sr;
         for(int i=ll;i>=;i-=(i&(-i)))nl[++lenl]=root[i];
         for(int i=rr;i>=;i-=(i&(-i)))nr[++lenr]=root[i];
         while(l<r)
             {
                 mid=(l+r)>>;
                sl=sr=;
                 for(int i=;i<=lenl;i++)sl+=t[t[nl[i]].ls].cnt;
                 for(int i=;i<=lenr;i++)sr+=t[t[nr[i]].ls].cnt;
                 if(sr-sl>=rank)
                     {
                         for(int i=;i<=lenl;i++)nl[i]=t[nl[i]].ls;
                         for(int i=;i<=lenr;i++)nr[i]=t[nr[i]].ls;
                         r=mid;
                     }
                 else
                     {
                         for(int i=;i<=lenl;i++)nl[i]=t[nl[i]].rs;
                         for(int i=;i<=lenr;i++)nr[i]=t[nr[i]].rs;
                         l=mid+;rank-=sr-sl;
                     }
             }
         return b[r];
     }
 void Clear()
     {
         tot=;num=;
         memset(root,,sizeof(root));
         memset(t,,sizeof(t));
     }
 void work()
     {
         Clear();
         for(int i=;i<=n;i++)scanf("%d",&a[i]),b[++num]=a[i];
         scanf("%d",&m);
         for(int i=;i<=m;i++)
             {
                 scanf("%d",&q[i].flag);
                 if(q[i].flag==){
                     scanf("%d%d%d",&q[i].x,&q[i].y,&q[i].k);
                 }
                 else{
                     scanf("%d%d",&q[i].x,&q[i].y);
                     b[++num]=q[i].y;
                 }
             }
         sort(b+,b+num+);
         for(int i=;i<=n;i++)bel[i]=getrank(a[i]);
         for(int i=;i<=n;i++)updata(i,bel[i],);
         for(int i=;i<=m;i++)
             {
                 if(q[i].flag==)
                     {
                         updata(q[i].x,bel[q[i].x],-);
                         updata(q[i].x,q[i].y=getrank(q[i].y),);
                         bel[q[i].x]=q[i].y;
                     }
                     else
                      printf("%d\n",query(q[i].x-,q[i].y,q[i].k));
             }
     }
 int main()
 {
     //freopen("pp.in","r",stdin);
     //freopen("pp.out","w",stdout);
     while(~scanf("%d",&n))
         work();
     return ;
 }